Need Help with solving a right triangle

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
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Need Help with solving a right triangle

Postby Blackhole252 » Fri Nov 13, 2009 10:46 pm

Triangle ABC is a right triangle, with B being the right angle and line CB being the ground. An obtuse triangle is inside triangle ABC. Let's call this obtuse triangle ADC.

The diagram should look something like this:

Suppose angle ACB=24.4 degrees and angle ADB=58.3 degrees, and CB=169 units.
AB should 106.5 units, but how do you solve this? Please explain the steps using sine, cosine and tangent ratios involving a right triangle.

An email would be appreciated if a solution is posted: [deleted for security purposes]

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Postby stapel_eliz » Sat Nov 14, 2009 1:08 am

You have a height h and a length x = |DB|.

You know that tan(24.4°) = h/(169 + x) and tan(58.3°) = h/x.

Multiplying through, you can get two equations in "h=". Set the other sides (the sides without the "h") equal, and solve for "x". :wink:

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